Question: Let $f$ be a vector-valued function defined by $f(t)=(4t^3+t,3\ln(t))$. Find $f'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(12t^2+1,\dfrac3t\right)$ (Choice B) B $\left(4t^2+1,-\dfrac3t\right)$ (Choice C) C $\left(t^4+\dfrac{t^2}2,3t\ln(t)-3t\right)$ (Choice D) D $4t^2+1+\dfrac3t$
Explanation: $f$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $f(t)=(4t^3+t,3\ln(t))$. Let's differentiate the first expression: $\begin{aligned}\dfrac{d}{dt}(4t^3+t)&=3\cdot4t^2+1 \\\\&=12t^2+1\end{aligned}$ Let's differentiate the second expression: $\dfrac{d}{dt}[3\ln(t)]=\dfrac3t$ Now let's put everything together: $\begin{aligned} f'(t)&=\left(\dfrac{d}{dt}(4t^3+t),\dfrac{d}{dt}[3\ln(t)]\right) \\\\ &=\left(12t^2+1,\dfrac3t\right) \end{aligned}$ In conclusion, $f'(t)=\left(12t^2+1,\dfrac3t\right)$.